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authorPrefetch2023-04-02 16:57:12 +0200
committerPrefetch2023-04-02 16:57:12 +0200
commita8d31faecc733fa4d63fde58ab98a5e9d11029c2 (patch)
treeb8d039b13e026fb68f0bed439a2cb73397c35981 /source/know/concept/blochs-theorem
parent9b9346d5e54244f3e2859c3f80e47f2de345a2ad (diff)
Improve knowledge base
Diffstat (limited to 'source/know/concept/blochs-theorem')
-rw-r--r--source/know/concept/blochs-theorem/index.md46
1 files changed, 16 insertions, 30 deletions
diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md
index 6f445f1..d7fcf90 100644
--- a/source/know/concept/blochs-theorem/index.md
+++ b/source/know/concept/blochs-theorem/index.md
@@ -17,85 +17,72 @@ take the following form,
where the function $$u(\vb{r})$$ is periodic on the same lattice,
i.e. $$u(\vb{r}) = u(\vb{r} + \vb{a})$$:
-$$
-\begin{aligned}
+$$\begin{aligned}
\boxed{
\psi(\vb{r}) = u(\vb{r}) e^{i \vb{k} \cdot \vb{r}}
}
-\end{aligned}
-$$
+\end{aligned}$$
In other words, in a periodic potential,
the solutions are simply plane waves with a periodic modulation,
known as **Bloch functions** or **Bloch states**.
-This is suprisingly easy to prove:
+This is surprisingly easy to prove:
if the Hamiltonian $$\hat{H}$$ is lattice-periodic,
then both $$\psi(\vb{r})$$ and $$\psi(\vb{r} + \vb{a})$$
are eigenstates with the same energy:
-$$
-\begin{aligned}
+$$\begin{aligned}
\hat{H} \psi(\vb{r}) = E \psi(\vb{r})
\qquad
\hat{H} \psi(\vb{r} + \vb{a}) = E \psi(\vb{r} + \vb{a})
-\end{aligned}
-$$
+\end{aligned}$$
Now define the unitary translation operator $$\hat{T}(\vb{a})$$ such that
$$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$$.
From the previous equation, we then know that:
-$$
-\begin{aligned}
+$$\begin{aligned}
\hat{H} \hat{T}(\vb{a}) \psi(\vb{r})
= E \hat{T}(\vb{a}) \psi(\vb{r})
= \hat{T}(\vb{a}) \big(E \psi(\vb{r})\big)
= \hat{T}(\vb{a}) \hat{H} \psi(\vb{r})
-\end{aligned}
-$$
+\end{aligned}$$
In other words, if $$\hat{H}$$ is lattice-periodic,
then it will commute with $$\hat{T}(\vb{a})$$,
i.e. $$[\hat{H}, \hat{T}(\vb{a})] = 0$$.
Consequently, $$\hat{H}$$ and $$\hat{T}(\vb{a})$$ must share eigenstates $$\psi(\vb{r})$$:
-$$
-\begin{aligned}
+$$\begin{aligned}
\hat{H} \:\psi(\vb{r}) = E \:\psi(\vb{r})
\qquad \qquad
\hat{T}(\vb{a}) \:\psi(\vb{r}) = \tau \:\psi(\vb{r})
-\end{aligned}
-$$
+\end{aligned}$$
Since $$\hat{T}$$ is unitary,
its eigenvalues $$\tau$$ must have the form $$e^{i \theta}$$, with $$\theta$$ real.
Therefore a translation by $$\vb{a}$$ causes a phase shift,
for some vector $$\vb{k}$$:
-$$
-\begin{aligned}
+$$\begin{aligned}
\psi(\vb{r} + \vb{a})
= \hat{T}(\vb{a}) \:\psi(\vb{r})
= e^{i \theta} \:\psi(\vb{r})
= e^{i \vb{k} \cdot \vb{a}} \:\psi(\vb{r})
-\end{aligned}
-$$
+\end{aligned}$$
Let us now define the following function,
keeping our arbitrary choice of $$\vb{k}$$:
-$$
-\begin{aligned}
+$$\begin{aligned}
u(\vb{r})
- = e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
-\end{aligned}
-$$
+ \equiv e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
+\end{aligned}$$
As it turns out, this function is guaranteed to be lattice-periodic for any $$\vb{k}$$:
-$$
-\begin{aligned}
+$$\begin{aligned}
u(\vb{r} + \vb{a})
&= e^{- i \vb{k} \cdot (\vb{r} + \vb{a})} \:\psi(\vb{r} + \vb{a})
\\
@@ -104,8 +91,7 @@ $$
&= e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
\\
&= u(\vb{r})
-\end{aligned}
-$$
+\end{aligned}$$
Then Bloch's theorem follows from
isolating the definition of $$u(\vb{r})$$ for $$\psi(\vb{r})$$.